sloof3: No, the point is: no matter WHAT input, you get the SAME output, for the SAME function definition.

sloof3

Capso: why would we get the same output

Capso

sloof3: That's the whole thing, isn't it?sloof3: You know your output, you know what the system is doing, you just don't know what you gave the system.

sloof3

I was thinking in terms of a system with an observable output and changing input

Capso

Your 'observability' defines the probabilities of what you MIGHT have given the system.

sloof3

Capso: yes

Capso

So, the OUTPUT needs to stay the same.

sloof3

If the output always stayed the same that wouldn't help us would it?

Capso

You can only CONSIDER one output.All the input values which would get that output.

sloof3

We don't know the inputs though

Capso

The thing is: you're not DETERMINING anything, you're simply getting a rating of a definition FROM an observation.Right, we don't know input.

sloof3

Originally I was only considering a system that had changing inputs.We would need at least n outputs for n inputs

Capso

Sorry, had some disruption here.No, your GOAL is to define 'observability' by knowing ONLY the following:(1) The number of inputs a function might take(2) The operation (definition) of the function(3) The output(s?) of the function.And the 'observability' will simply be you *ability to define the inputs to obtain the output(s?)*.

sloof3

define the inputs given the outputs

Capso

That's vague.What I've stated is clearer.The reason I include the '(s?)' in 'output(s?)' is that I'm still considering whether we should take into account > 1 output.You need certain inputs to GET those outputs.Okay, suppose:f(x,y) = x + y; -- Definitionf(x,y) = 5f(x,y) = 6Etc.